# Convergent Real Sequence/Examples/Arithmetic Mean of Previous 2 Terms

## Example of Convergent Real Sequence

Let $\sequence {x_n}_{n \mathop \in \N_{>0} }$ be the sequence in $\R$ defined as:

$x_n = \begin {cases} a & : n = 1 \\ b & : n = 2 \\ \dfrac {x_{n - 1} + x_{n - 2} } 2 & : n > 2 \end {cases}$

That is, beyond the first $2$ terms, each term is the arithmetic mean of the previous $2$ terms.

Then $\sequence {x_n}$ converges.

## Proof

 $\displaystyle x_{n + 2} - x_{n + 1}$ $=$ $\displaystyle \dfrac {x_{n + 1} + x_n} 2 - x_{n + 1}$ $\displaystyle$ $=$ $\displaystyle \dfrac {x_n - x_{n + 1} } 2$ $\displaystyle \leadsto \ \$ $\displaystyle \size {x_{n + 2} - x_{n + 1} }$ $=$ $\displaystyle \dfrac {\size {x_n - x_{n + 1} } } 2$ $\displaystyle$ $=$ $\displaystyle \dfrac {\size {x_n - x_{n - 1} } } {2^2}$ $\displaystyle$ $=$ $\displaystyle \ldots$ $\displaystyle$ $=$ $\displaystyle \dfrac {\size {x_2 - x_1} } {2^n}$ $\displaystyle$ $=$ $\displaystyle \dfrac {\size {b - a} } {2^n}$

Let $n > m$.

Then:

 $\displaystyle \size {x_n - x_m}$ $=$ $\displaystyle \size {\paren {x_n - x_{n - 1} } + \paren {x_{n - 1} - x_{n - 2} } + \dotsb + \paren {x_{m + 1} - x_m} }$ $\displaystyle$ $\le$ $\displaystyle \size {x_n - x_{n - 1} } + \size {x_{n - 1} - x_{n - 2} } + \dotsb + \size {x_{m + 1} - x_m}$ Triangle Inequality for Real Numbers $\displaystyle$ $\le$ $\displaystyle \paren {\dfrac 1 {2^{n - 2} } + \dfrac 1 {2^{n - 3} } + \dotsb + \dfrac 1 {2^{m - 1} } } \size {b - a}$ $\displaystyle$ $=$ $\displaystyle \dfrac 1 {2^{m - 1} } \paren {1 + \dfrac 1 2 + \dfrac 1 {2^2} + \dotsb + \dfrac 1 {2^{n - m - 1} } } \size {b - a}$ $\displaystyle$ $=$ $\displaystyle \dfrac 1 {2^{m - 1} } \paren {\dfrac {1 - \paren {\frac 1 2}^{n - m} } {1 - \frac 1 2} } \size {b - a}$ Sum of Geometric Sequence $\displaystyle$ $\le$ $\displaystyle \dfrac 1 {2^{m - 2} } \size {b - a}$

Let $\epsilon \in \R_{>0}$ be given.

Let $N$ be sufficiently large that:

$\dfrac 1 {2^{N - 2} } \size {b - a} < \epsilon$

Then:

$\forall n > N, m > N: \size {x_n - x_m} \le \dfrac 1 {2^{N - 2} } \size {b - a} < \epsilon$

Hence it is seen that $\sequence {x_n}$ is a Cauchy sequence.

Hence the result, by Cauchy Sequence Converges on Real Number Line.

$\blacksquare$